1. Given data

• Voltage amplitude: $V_0 = 40\,\text{V}$
• Frequency: $f = 4\,\text{kHz} = 4 \times 10^3\,\text{Hz}$
• Capacitance: $C = 12\,\mu\text{F} = 12 \times 10^{-6}\,\text{F}$

We need the maximum displacement current through the capacitor.

2. Use the capacitor current relation

For an AC voltage across a capacitor,

$i = C\frac{dv}{dt}$

If

$v = V_0 \sin \omega t,$

then

$i = C\omega V_0 \cos \omega t$

So the maximum current is

$I_0 = \omega C V_0$

where

$\omega = 2\pi f$

3. Calculate angular frequency

$\omega = 2\pi (4 \times 10^3) = 8\pi \times 10^3\,\text{rad/s}$

4. Calculate maximum current

$I_0 = \omega C V_0$

$I_0 = (8\pi \times 10^3)(12 \times 10^{-6})(40)$

Now simplify:

$10^3 \times 10^{-6} = 10^{-3}$

So,

$I_0 = 8\pi \times 12 \times 40 \times 10^{-3}$

$I_0 = 3.84\pi$

Using $\pi \approx 3.14$,

$I_0 \approx 3.84 \times 3.14 \approx 12.06\,\text{A}$

5. Nearest option

$I_0 \approx 12\,\text{A}$

So the correct option is D.

6. Comparison with stored answer

Stored correct answer: D

My derived answer: D

They agree.